1 kg of 0.75 molal aqueous — JEE Mains Chemistry Past Papers Chemistry Question

💡 Solution & Explanation

Let mass of water initially present = x g Mass of sucrose = (1000 – x) g Number of moles sucrose = x 1000 Molality (M) = / x x 1000 = 0.75 2565.5 x = 106 – 1000x x = 795.86 g no. of moles of sucrose = 0.5969 Tƒ = ikƒm 4 = 1 × 1.86 × g ) a . ( .  × 1000 Ice separated = 518.3 g